-13t^2+96t+4=0

Solution for -13t^2+96t+4=0 equation:

-13t^2+96t+4=0

a = -13; b = 96; c = +4;
Δ = b2-4ac
Δ = 962-4·(-13)·4
Δ = 9424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9424}=\sqrt{16*589}=\sqrt{16}*\sqrt{589}=4\sqrt{589}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-4\sqrt{589}}{2*-13}=\frac{-96-4\sqrt{589}}{-26}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+4\sqrt{589}}{2*-13}=\frac{-96+4\sqrt{589}}{-26}
$